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Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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We've seen how a chemical equilibrium can be taken out of equilibrium by adding a reagent, in other words, by increasing the amount of a reactant. For instance, we saw that the system had to shift in such a way to use up some of that added reactant, in order to re-establish its equilibrium. Let's ask a different question now. Let's ask how an equilibrium will respond to a change in volume or pressure, but with the situation that we're not adding material to the system this time. We're simply going to change the volume of the container or change the pressure of the overall system. And again, we want to ask how the system will respond to that.
So let's take an example of N[2]O[4], equilibrium with NO[2]. We've talked about this equilibrium before. And let's suppose that we start out with this system at equilibrium at 1 atmosphere. And I want to know what will happen to that equilibrium if I change the pressure. Specifically, I want to know what happens if I double the pressure. Well, I have two different ways that I can double that pressure. I can either add argon, some inert substance that is not involved in the reaction at all, but simply plays the role of doubling the pressure, simply because I'm adding an inert substance into the same container. So my total pressure increases. Or I can double the pressure by lowering the volume. Now, how would I do that? Well, I could have a piston in a cylinder, for instance, with this equilibrium in that and I could change the volume of my container by lowering down the piston. And, in doing so, I wouldn't be affecting the amount of stuff that I had in my equilibrium. I'm not adding NO[2] or N[2]O[4] in either of these situations, but I am affecting the pressure. And I can ask how will this equilibrium respond? And, specifically, I just want to know will I be taken out of equilibrium? That's the first question, really, I should ask. And, if I am taken out of equilibrium, what way does it have to respond to get back to equilibrium? So let's, first of all, consider the case of adding an inert material, where we increase total pressure, but not changing the volume.
Now, let's suppose that, at 1 atmosphere of pressure, I'm at a temperature such that my equilibrium constant is .13. And the particular pressures that I have here at equilibrium initially are .3 atmospheres for the NO[2] and .7 atmospheres for the N[2]O[4]. So my ratio of products over reactants is going to be equal to my equilibrium constant we know. Now, I go ahead and I increase my total pressure by adding argon to raise the pressure up to two atmospheres. What happens to the system? What happens, in particular to that equilibrium? Well, by adding argon, what have I indeed done to the partial pressures? Not a thing. Think about that for a moment. When I add argon, if I were a molecule of NO[2], I'm essentially unaware that another gas has been introduced. I know that I have the same volume to wonder around in. I have no recollection of a pressure on the walls. Now, the pressure on the walls changes, because there's more stuff in there, but the volume taken up by the argon is insignificant compared to the total volume. So the pressure experienced by the NO[2] molecules doesn't change at all. In other words, simply put, the partial pressures are not affected by the addition of other materials, like argon. So the ratio of NO[2] and N[2]O[4], that has not changed at all, because the partial pressures haven't changed. In other words, we've never left equilibrium. By doubling the pressure, the system has no need to respond, because it's still in equilibrium. We might interpret Le Chatelier's principle too literally if we say, "Well, you apply a stress and the system needs to shift to relieve the stress. If the pressure increases the stress, the system could relieve some of that pressure, by running the reaction from right to left," and that would make intuitive sense. But, unfortunately, it's not correct, because, in this case, although the total pressure changed, the partial pressures didn't. And the partial pressures are what show up in the equilibrium expression, not the total pressure. So no change whatsoever if we add an inert substance, as long as we keep our volume the same.
Now, let's look at the other situation. Let's suppose that, instead of adding the argon, we lowered the volume of the container. So, in this case, what we're going to do is, again, imagine that we've got a piston and cylinder and we shrink the volume to half it's original value. Now, by Boyle's law, PV is constant, so we go from 1 atmosphere to two atmospheres, as long as we're not changing temperature. And so, now what happens to the partial pressure? We'll we've reduced the volume that the NO[2], for instance, has to move around in. So it's pressure, in this case, has to double, because we're lowered the volume by a factor of two. Likewise, the partial pressure of the N[2]O[4] must double. And again, remember this is the instant after changing volume, before any changes have occurred. So those numbers both double. But that ratio now, because of the squared term, has not stayed the same. This value now is .26, again because it's NO[2] squared over the N[2]O[4]. So this value is larger than it was previously, and so q is now larger than k. So the reaction must shift to the left, in order to lower the amount of NO[2] and raise the amount of N[2]O[4]. So to restore equilibrium now, the system must move from right to left, making fewer molecules, in line with Le Chatelier's principle that if you make fewer moles of gas, you reduce that pressure. But again, this time the stress is very real. It's a change of partial pressures now, and that's why the system was taken out of equilibrium and had to respond by changing the overall ratio of reactants and products to bring us into this new equilibrium, to make q, again, equal to kp.
Now, let's look at a very similar idea conceptually, except that, instead of talking about gases, let's talk about something dissolved in solution. Let's look at an organic reaction. This is an example, again, of something called a cyclo addition reaction. The only important thing for you to understand right now is that it's two molecules, the cyclopentadiene, coming together to make 1 molecule, a dimer, if you will, of the original material. This is an equilibrium in solution. Let's say we're at a given temperature, where we've got equilibrium established in some organic solvent, maybe acetonitrile. And what we're going to ask is what happens when we double the volume of the solution? How is this equilibrium affected, if at all? We haven't changed, again, the amounts of either of these things. All we've changed is the volume in which these materials are free to move around in. So, in that sense, it's very, very analogous to the last example we saw. Instead of changing the volume of the container, we change the volume of the solvent, which effectively is the container, as far as these guys are concerned. They `re free to wander around in that solvent, but not in the whole container. So what happens to this equilibrium? Well, Le Chatelier's principle, to apply that here, we'd say that the stress is the increase in volume. The system would respond to decrease that stress by filling the extra volume with more molecules. How does it do it? It goes from the right to the left, because that creates more molecules. And so, we'd expect a shift of the reaction in this direction to establish a new equilibrium. Well, let's see if that's actually true.
With starting concentrations at the original equilibrium of .32 for the dimer and .0046 for the cyclopentadiene, that originally has a value of 1.5 times 10^4. Now notice again, this is kc, not kp, because we talking now in terms of concentrations. And, if I double the volume, what do I do to my concentrations? Well, I haven't changed my moles, but I've changed my volume, so my concentration has to decrease by two. Again, it's moles per liter. So we've changed our volume by two, so our concentrations have dropped by two. Plugging those in for the reaction quotient, like we did before, we find that the value now is twice as large as it was before. So what we've done is taken the system out of equilibrium, q no longer equals k, and so it must respond. In order to get back to equilibrium, q has to decrease and that means we have to lower the numerator or increase the denominator. Physically, that means the system has got to run from the right to the left. It has to go in the back direction to make more of the monomer, more of the cyclopentadiene, just like what we predicted with Le Chatelier's principle.
Well finally, what about a system where there aren't different numbers of moles on the two sides? We have two moles overall here, one mole overall of stuff here. And the system could improve its situation simply by making more moles of stuff. But what if those coefficients turned out to be the same? So let's go back, once again, to the H[2] and I[2] example. In this case, we have equal numbers of moles of gas on the left and on the right. Now, in this case, we could ask what happens if we double the volume, let's say, similar to that last problem. We double the volume - this time, we're talking about a gas, but it's the same idea. We'll double the volume. What happens to the equilibrium? Well, in this case, there's nothing the reaction can do to alter its situation. There are equal numbers of moles on both sides. So, in fact, we would predict that the system would do absolutely nothing. And, in fact, that's true. I challenge you to do exactly what we did earlier, when we changed the volume, and solve for partial pressures of all of these things. What you'll find is, by doubling the volume, the partial pressure of these guys all goes down, but, if you plug those values into the reaction quotient, you'll find that the value still equals the equilibrium constant, telling you that you're still at equilibrium. The system does not need to respond to that change in volume, and the reason is because there are equal numbers of moles on the left and on the right side.
So, what have we learned? Le Chatelier's principle is a very powerful tool to allow us to predict what direction a system must shift, how it must respond, in order to restore an equilibrium. And also, very important as part of that, it tells us whether we've left equilibrium at all. So remember that the important thing is what happens to concentrations or partial pressures, not what happens to the total pressure of the system. Those, after all, are the components that are in the equilibrium ratio. And so, it's only when those values are affected that we are on the lookout for q no longer being equal to k, and we have to consider the reaction shifting, in order to restore that equilibrium.
Chemical Equilibrium
Shifting Chemical Equilibrium
The Effect of Pressure and Volume on Equilibrium Page [1 of 2]

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